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3 years ago

How is capacitance related to the distance between the plates of a capacitor?

Physics

1 answer:

svetlana [45]3 years ago

5 0

Answer:C

Explanation: I studied, and C is correct

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Help with i) and ii) pls >_

alina1380 [7]

Answer:

4 it the corrcert anser

Explanation:

4 0

2 years ago

Read 2 more answers

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

ololo11 [35]

Answer:

0.027 J

Explanation:

The formula of the potential energy in electrostatic (U) is

$U = q \times E \times d$

The values given in the question are

Charge : q = $4.5 \times 10^{-5} C$

Electric Field strength : E = $2.0 \times 10^{4} N/C$

Distance : d is 0.030 m

Insert in the formula , will give us

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

Further solving it

$U = 0.027 J$

Which is the required answer.

Thanks

5 0

2 years ago

A brass washer has an outside diameter of 4.50 cm with a hole of diameter 1.25 cm and is 1.50 mm thick. The density of brass is

-Dominant- [34]

We will find the mass from
mass = density x volume
We are told the density and must find the volume from the dimensions given
the volume of the washer will be the area x thickness (remembering to convert all measurements to meters)
if the washer had no hole, its area would be pi (0.0225m)^2 (remember to convert to meters and to use radius)
the area of the hole is pi(0.00625m)^2
so the area of the washer is pi[(0.0225m)^2 - (0.00625m)^2] = 1.5x10^-3 m
the volume of the washer is 1.5x10^-3 m x 1.5x10^-3 m = 2.25x10^-6 m^3 (the thickness of the washer is 1.5 mm = 1.5x10^-3m)
thus, the mass of the washer = 8598kg/m^3 x 2.25x10^-6m^3 = 0.0189kg = 18.9 grams

6 0

2 years ago

What is the momentum of a 3 kg bowling ball moving at 3 m/s?

Nataly [62]

Explanation:

p denotes momentum
m denotes mass
v denotes velocity

→ p = 3 kg × 3 m/s

→ p = 9 kg.m/s

Option D is correct.

5 0

2 years ago

A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.

sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0

2 years ago